Divide the following complex numbers. $ \dfrac{-1+32i}{4-5i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${4+5i}$ $ \dfrac{-1+32i}{4-5i} = \dfrac{-1+32i}{4-5i} \cdot \dfrac{{4+5i}}{{4+5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-1+32i) \cdot (4+5i)} {(4-5i) \cdot (4+5i)} = \dfrac{(-1+32i) \cdot (4+5i)} {4^2 - (-5i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-1+32i) \cdot (4+5i)} {(4)^2 - (-5i)^2} = $ $ \dfrac{(-1+32i) \cdot (4+5i)} {16 + 25} = $ $ \dfrac{(-1+32i) \cdot (4+5i)} {41} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-1+32i}) \cdot ({4+5i})} {41} = $ $ \dfrac{{-1} \cdot {4} + {32} \cdot {4 i} + {-1} \cdot {5 i} + {32} \cdot {5 i^2}} {41} $ Evaluate each product of two numbers. $ \dfrac{-4 + 128i - 5i + 160 i^2} {41} $ Finally, simplify the fraction. $ \dfrac{-4 + 128i - 5i - 160} {41} = \dfrac{-164 + 123i} {41} = -4+3i $